\newproblem{lay:5_1_19}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.1.19}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	For $A=\begin{pmatrix}1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3\end{pmatrix}$, find one eigenvalue, with no calculation. Justify your answer.
}{
  % Solution
	The determinant of $A$ is zero because it has duplicated rows. On the other hand, the determinant is the product of the matrix eigenvalues, so at least one of
	the eigenvalues of $A$ must be zero.
}
\useproblem{lay:5_1_19}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
